Important Questions Class 9 Maths Chapter 10

Important Questions Class 9 Mathematics Chapter 10- Circles

One subject that calls for a lot of focus from students in Mathematics. Success might result from a successful combination of a strategic approach and a lot of practice. This subject can help students achieve excellent grades. In Chapter 10, students will learn about circles, various terms that are linked to them, and some other characteristics of circles.

Students’ dedication, along with Important Questions Class 9 Mathematics Chapter 10 developed by Extramarks, ensures their success. Mathematics from the ninth grade serves as a basis leading up to board exams. Prior to entering Class 10, students must possess clearer concepts. These Important Questions Class 9 Mathematics Chapter 10 serve as the foundation for the final exam and give students greater confidence as they approach their tests.

Students can complete these Chapter 10 Class 9 Mathematics important questions just before their final exams. Years of experience from the Extramarks team in both teaching and learning might help students  clear up their most pressing questions. Important Questions Class 9 Mathematics Chapter 10 are only released after taking into account the latest syllabus. Just prior to the test, students can readily go through them. Students can quickly clear up their confusion while working through the exercise problems in the NCERT book with the aid of these Important Questions Class 9 Mathematics Chapter 10.

The Important Questions Class 9 Mathematics Chapter 10 is an enduring solution for all exams, whether they are at the state or national level. It may serve as a convenient solution for students. Making sure that students don’t miss any topics is the key priority. 

Most students are unable to finish every chapter of their syllabus. So they can simply solve problems with the Important Questions Class 9 Mathematics Chapter 10. This assists students in resolving difficult mathematical problems.

Important Questions Class 9 Mathematics Chapter 10 is prepared based on the NCERT textbook and CBSE curriculum. Our topic specialists have developed these questions in order to help students do well on their final exams. Additionally, some more resources are provided by Extramarks for better practice and revision.

Important Questions Class 9 Mathematics Chapter 10 – With Solutions

One of the most crucial chapters in Class 9 is Circles and its ideas are also employed in Class 10. This chapter holds significant weightage in the final test. In order to have a good practice, it is advised that students read the chapter thoroughly and practise all of the questions in the textbook using the NCERT solutions and examples.

These are a few questions and their answers from our list of Mathematics Class 9 Chapter 10 important questions.

Question: 1. AD is the diameter of the circle, and AB is the chord. If AD = 34 cm, AB = 30 cm, the distance AB from the centre of the given circle is :

(A) 4 cm 

(B) 15 cm

(C) 17 cm

(D) 8 cm

Solution 1:

(D) 8 cm

Explanation:

Diameter of the given circle = d = AD = 34 cm

∴ Radius of circle = r = d/2 = AO = 17 cm

Length of the chord AB = 30 cm

Since the line being drawn through the centre of the circle bisecting the chord is perpendicular to the chord. Thus AOP is a right-angled triangle with L being the bisector of AB.

∴ AP = 1/2(AB) = 15 cm

In the right-angled triangle AOB, using Pythagoras theorem, we get:

(AO)2 = (OP)2 + (AP)2

⇒ (17)2 = (OP)2 + (15)2

⇒ (OP)2 = (17)2 – (15)2

⇒ (OP)2 = 289 – 225

⇒ (OP)2 = 64

Taking square roots on both sides:

⇒ (OP) = 8

Thus, the distance of AB from the center of the circle is 8 cm.

Question:2. In figure given below, given OA = 5 cm, AB = 8 cm and OD which is perpendicular to AB, then CD will be equal to:

(A) 2 cm

(B) 5 cm

(C) 4 cm

(D)  3 cm

Solution 2:

(A) 2 cm

Explanation:

Given:

Radius of circle = r = AO = 5 cm

Length of chord AB = 8 cm

AOC is a right-angled triangle with C as the bisector of AB because the line drawn from the centre of the circle, such as to bisect a chord, is perpendicular to the chord.

∴ AC = ½ (AB) = 8/2 = 4 cm

In the right-angled triangle AOC, using Pythagoras theorem, we have:

(AO)2 = (OC)2 + (AC)2

⇒ (5)2 = (OC)2 + (4)2

⇒ (OC)2 = (5)2 – (4)2

⇒ (OC)2 = 25 – 16

⇒ (OC)2 = 9

Taking square roots on both sides:

⇒ (OC) = 3

∴ The distance of AC from the centre of circle is 3 cm.

Now, OD is Radius of the circle, ∴ OD = 5 cm

CD = OD – OC

CD = 5 – 3

CD = 2

Therefore, CD = 2 cm

Question:3. Given AB = 12 cm, BC = 16 cm and AB which is perpendicular to BC, then the radius of the circle which passes through the points A, B and C is :

(A) 6 cm

(B) 12 cm

(C) 10 cm

(D)  8 cm

Solution 3:

(C) 10 cm

Explanation:

As per the question,

AB = 12 cm, BC = 16 cm, AB ⊥ BC.

Thus,

AC is the diameter of a given circle passing through points A, B and C.

Now, from the figure,

We get,

ABC is a right-angled triangle.

Using Pythagoras theorem:

(AC)2 = (CB)2 + (AB)2

⇒ (AC)2 = (16)2 + (12)2

⇒ (AC)2 = 256 + 144

⇒ (AC)2 = 400

Taking square root on LHS and RHS,

We get,

(AC) = 20

Diameter of circle = 20 cm

Thus, Radius of circle = Diameter/2

= 20/2

= 10 cm

Thus, radius of the circle = 10 cm

Question:4. In figure given below, when ∠ABC = 20º, then ∠AOC will be equal to:

(A) 60º

(B) 40º

(C) 10º

(D) 20º

Solution 4:

 (B) 40º

Explanation:

As per the question,

∠ABC = 20°

Now we know,

“The angle subtended by any arc at the center of the circle is twice such the angle subtended by it at the other remaining part of the circle.”

According to the theorem, we have,

∠AOC = 2 × ∠ABC

= 2 × 20°

= 40°

Therefore, ∠AOC = 40°

Hence, option B is the correct answer.

Question:5. In the figure given below, when AOB is the diameter of the circle while AC = BC, then ∠CAB is equal to:

(A) 90º

(B) 60º

(C) 30º

(D) 45º

Solution 5:  (D) 45º

Explanation:

As per the question,

We have,

Diameter of circle = AOB

AC = BC

Since the opposite angles to the equal sides are equal

∠ABC = ∠BAC

We suppose, ∠ABC = ∠BAC = x

And diameter subtends at a right angle to the circle,

∠ACB = 90°

We also know,

Using the angle sum property of a triangle, sum of all angles of a triangle = 180°.

∠CAB + ∠ABC + ∠ACB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 90°

⇒ x = 45°

∠CAB = ∠ ABC = 45°

Hence, option D is the correct answer.

Question:6. An angle in the semi-circle is

(a) Right angle

(b) 1800

(c)3600

(d) None of these

Solution 6: (a) Right angle

Explanation: When an angle is inscribed in a semi-circle, it will be half the measure of a semi-circle (180∘), therefore measuring 90∘.

Question:7. Fill in the gaps:

(i) The centre of the circle lies in the__________ of a circle.

Answer: Interior

(ii) A point, the distance of which from the centre of the circle is greater than its radius, lying in_______________ of the circle.

Answer: Exterior

(iii) The longest chord of circle is the _______________ of the circle.

Answer: Diameter

(iv) An arc is a _______________ when its ends are also the ends of a diameter.

Answer: Semi-circle

(v) Segment of circle is the region between an arc and the _______________ of the circle.

Answer: Chord

Question:8. Write True/False:

(i) Line segment which is joining the centre to any other point on the circle is the radius of the circle.

Answer: True

(ii) A circle has a finite number of equal chords.

Answer: False

(iii) When a circle is divided into three equal arcs, then each is a major arc.

Answer: False

(iv) A chord is twice as long as its radius is the diameter of the circle.

Answer: True

(v) Sector is a region between the chord and its corresponding arc.

Answer: False

(vi) A circle is a plane figure.

Answer: True

Question:9. Two chords, such as AB and CD, of a circle are both at distances 4 cm from the centre. Then AB = CD.

Solution 9:

 It is true.

Justification:

Given that AB and AC are chords which are at a distance of 4 cm from the centre of a circle.

We know chords that are equidistant from the centre of the circle are equal in length,

We have AB = CD.

Question:10. Two chords AB and AC of a circle with its centre O are on the opposite sides of OA. Then ∠OAB = ∠OAC.

Solution 10:

 It is false.

Justification:

We suppose AB and AC be the chord of a circle with its centre O on the opposite side of OA.

Considering the triangles AOC and AOB:

AO = AO (Common side in both the triangles)

OB = OC (Both OB and OC are the radius of circle)

But we can’t show that either the third side of both the triangles are equal or any of the angles is equal. Therefore ΔAOB is not congruent to ΔAOC.

∴ ∠OAB ≠ ∠OAC.

Question:11. Two congruent circles with their centres O and O’ intersect at  A and B. Then ∠AOB = ∠AO’B.

Solution 11:

 It is true

Justification:

Equal chords of the congruent circles subtend equal angles at the respective centre.

Hence, the given statement is true.

Question:12. Through the three collinear points, a circle can be drawn.

Solution 12:

 It is False

Justification:

A circle through any two points cannot pass through a point which is collinear to these two points.

Question:13. A circle of 3 cm radius can be drawn through two points, A and B, such that AB = 6 cm.

Solution 13:

 It is True

Justification:

As per the question,

Radius of the circle = 3 cm

Diameter of the circle = 2 × r

= 2 × 3 cm

= 6 cm

Now,

we have from the question,

AB = 6 cm

So, AB will be the diameter.

Question:14. If arcs AXB and CYD of any circle are congruent, find the ratio of AB and CD.

Solution 14:

As per the question,

We have,

AXB ≅ CYD.

We know,

If two arcs of any circle are congruent, then their corresponding arcs are also equal.

So, we have chord AB = chord CD.

Hence, we get,

AB/CD = 1

AB/CD = 1/1

AB : CD = 1:1

Question:15. If the perpendicular bisector of chord AB of the circle PXAQBY intersects the circle at points P and Q, prove that arc PXA ∠ Arc PYB.

Solution 15:

As per the question,

We have,

PQ is a perpendicular bisect of AB,

Thus, we get,

AM = BM …equation (1)

In △APM and △BPM,

From equation (1),

AM = BM

∠AMP = ∠BMP = 90o

PM = PM [ as Common side]

Thus, △APM ≅ △BPM [By SAS congruence rule]

So, AP = BP [CPCT]

Hence, arc PXA ≅ Arc PYB

Therefore, if two chords of a circle are equal, then their corresponding arcs are congruent.

Question:16. A, B, and C are the three points on any circle. Prove that perpendicular bisectors of AB, BC, and CA are concurrent.

Solution 16:

As per the question,

Three non-collinear points, A, B, and C, are on the circle.

To prove: The perpendicular bisectors of AB, BC and CA are concurrent.

Construction: We join AB, BC and CA.

We draw:

ST, which is the perpendicular bisector of AB,

PM, which is the perpendicular bisector of BC

And, QR which is the perpendicular bisector of CA

Since points A, B and C are not collinear, ST, PM, and QR are not parallel and will intersect.

Proof:

O lies on ST, the perpendicular bisector of AB

OA = OB … eq(1)

Likewise, O lies on PM, the perpendicular bisector of BC

OB = OC … eq(2)

And, O lies on QR, the perpendicular bisector of CA

OC = OA … eq(3)

From equations (1), (2) and (3),

OA = OB = OC

Let OA = OB = OC = r

Drawing circle, with its centre O and radius r, passing through points A, B and C.

So, O is the only point which is equidistant from points A, B and C.

Thus, the perpendicular bisectors of AB, BC and CA are concurrent.

Question:17. AB and AC are any two equal chords of the circle. Prove that bisector of the angle BAC will pass through the centre of the circle.

Solution 17:

As per the question,

We have,

AB and AC are two equal chords with centre O.

AM is the bisector of ∠BAC.

To prove: AM passes through O.

Construction: Join BC.

Let AM intersect BC at P.

Proof: In DBAP and DCAP

AB = AC [Given]

∠BAP = ∠CAP [Given]

And AP = BP [Common side]

△BAP ≅ △CAP [By SAS]

∠BPA = ∠CPA [CPCT]

We know that,

CP = PB

But, as ∠BPA and ∠CPA are linear pair angles,

We have,

∠BPA+∠CPA =180o

∠BPA = ∠CPA = 90o

So, AP is a perpendicular bisector of the chord BC and will pass through the centre O on being produced.

Therefore, AM passes through O.

Question:18. If a line segment joining the mid-points of any two chords of a circle passes through the centre of circle, prove that the two chords are parallel.

Solution 18:

We take AB and CD to be the two chords of circle with center O.

Let L and M be the midpoint of AB

 and CD, respectively.

Let PQ be the line that passes through these midpoints and also the centre of the circle.

Then, PQ is the diameter of the circle.

We know,

The line joining the centre to the midpoint of a chord is always perpendicular to the chord.

Since M is the midpoint of CD,

We have OM ⊥ CD

⇒ OMD = 90°

Similarly, L is the midpoint of AB,

OL ⊥ AB

⇒ OLA = 90°

But, we know,

∠OLA and ∠OMD are alternate angles.

So, AB ∥ CD.

Hence, proved.

Question 19. ABCD is a quadrilateral where A is the centre of the circle which passes through points B, C, and D. Prove that ∠CBD + ∠CDB = ½ ∠BAD

Solution 19:

As per the question,

We have,

A quadrilateral ABCD where A is the centre of the circle, which passes through the points B, C and D.

Construction:

We join CA and BD.

We know,

An arc’s angle subtended at the centre of a circle is twice as large as its angle at any other point in the remaining portion of the circle.

So,

The arc DC which subtends ∠DAC at the centre and ∠CAB at point B in the remaining part of the circle,

We get

∠DAC = 2∠CBD …eq(1)

Similarly,

The arc BC, which subtends ∠CAB at the centre and ∠CDB at point D in the remaining part of the circle,

We get,

∠CAB = 2∠CDB …(2)

From equations (1) and (2),

We have:

DAC + ∠CAB = 2∠CDB + 2∠CBD

⇒ ∠BAD = 2(∠CDB + ∠CBD)

⇒ (∠CDB + ∠CBD) = ½ (∠BAD)

Question 20:  O is the circumcentre of the triangle ABC, while D is the mid-point of the base BC. Prove that ∠BOD = ∠A.

Solution 20:

As per the question,

We have,

O is the circumcenter of triangle ABC and D as the midpoint of BC.

To prove: ∠BOD = ∠A

Construction: We join OB and OC.

In ΔOBD and ΔCD:

OD = OD (common for both)

DB = DC (D is midpoint of BC)

OB = OC (Radius of the circle)

Using SSS congruence rule,

We get,

ΔOBD ≅ ΔOCD.

∠BOD = ∠COD (By CPCT)

Let ∠BOD = ∠COD = x

We know,

An arc’s angle subtended at the centre of a circle is twice as large as its angle at any other point in the remaining portion of the circle.

Thus, we have,

2∠BAC = ∠BOC

⇒ 2∠BAC = ∠BOD + ∠DOC

⇒ 2∠BAC = x + x

⇒ 2∠BAC = 2x

⇒ ∠BAC = x

⇒ ∠BAC = ∠BOD

Hence, proved.

Question 21: On the common hypotenuse AB, two right triangles, ACB and ADB, are situated on opposite sides. Prove that ∠BAC = ∠BDC.

Solution 21:

As per the question,

We have,

ACB and ADB are two right triangles.

To Prove: ∠BAC = ∠BDC

We know,

ACB and ADB are right-angled triangles,

Then,

∠C + ∠D = 90° + 90°

∠C + ∠D = 180°

Thus ADBC is a cyclic quadrilateral as the sum of the opposite angles of a cyclic quadrilateral = 180°

We also have,

∠BAC and ∠BDC lying in the same segment BC and angles in the same segment of a circle are equal.

∴ ∠BAC = ∠BDC.

Hence Proved.

Question 22: Two chords AB and AC of the circle subtend angles which are equal to 90º and 150º, respectively, at the centre. Find ∠BAC when AB and AC lie on opposite sides of the centre.

Solution 22:

As per the question,

We have,

In ΔAOB,

OA = OB (Radius of the circle)

Since the opposite angles to equal sides are equal, we get,

∠OBA = ∠OAB

We know,

According to the angle sum property, the sum of all angles of a triangle = 180o

By angle sum property in ΔAOB, we get,

∠OAB + ∠AOB +∠OBA = 180°

⇒ ∠OAB +90° + ∠OAB = 180°

⇒ 2∠OAB = 180° – 90°

⇒ 2∠OAB = 90°

⇒ ∠OAB = 45°

Now, in ΔAOC,

OA = OC (radius of circle)

Since opposite angles to equal sides are equal

∴ ∠OCA = ∠OAC

By the angle sum property in ΔAOB, sum of all angles of the triangle is 180°; we have:

∠OAC + ∠AOC +∠OCA = 180°

⇒ ∠OAC +150° + ∠OAC = 180°

⇒ 2∠OAC = 180° – 150°

⇒ 2∠OAC = 30°

⇒ ∠OAC = 15°

Now, ∠BAC = ∠OAB + ∠OAC

= 45° + 15°

= 60°

∴ ∠BAC = 60°

Question 23:  If BM and CN are perpendiculars, such as drawn on sides AC and AB of triangle ABC, prove that the points B, C, M, and N are concyclic.

Solution 23:

As per the question,

BM and CN are perpendiculars drawn on the sides AC and AB of triangle ABC.

So, we have,

∠BMC = ∠BNC = 90o

We know,

The four points are concyclic if a line segment connecting two points subtends equal angles on the same side of the line carrying the segment.

Considering the question,

B, C, M, and N are concyclic because BC connects the two points B and C that form the equal angles BMC and BNC at M and N on the same side BC containing the segment.

Thus, we get that,

B, C, M and N are concyclic.

Question 24:  If the two equal chords of a circle intersect, prove that parts of one chord would be separately equal to the parts of the other chord.

Solution 24:

As per the question,

AB and CD are the two equal chords of a circle with centre O, intersecting each other at M.

To prove:

(i) MB = MC and

(ii) AM = MD

Proof:

AB is any chord and OE perpendicular to AB from the centre O,

Since the perpendicular from the centre O  to a chord bisect the chord

We get,

AE = ½ AB

Similarly,

FD = ½ CD

Given that,

AB = CD⇒ ½ AB = ½ CD

So, AE = FD … (1)

Since the equal chords are equidistant from the centre,

And AB = CD

So, OE = OF

Now, as proven earlier, in right triangles MOE and MOF,

hyp. OE = hyp. OF [Common side]

OM = OM

ΔMOE ≅ ΔMOF

ME = MF … eq.(2)

Subtracting equations eq.(2) from eq(1), we get

AE – ME = FD – MF

⇒ AM = MD [Proved part (ii)]

Again, AB = CD [Given]

Also AM = MD [Proved]

AB – AM = CD – MD [i.e., equals subtracted from equal]

Therefore, MB = MC [Proved part (i)]

Question 25 If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution 25:

As per the question,

We have,

ABCD is a trapezium where AD||BC

Non-parallel sides AB and DC of trapezium ABCD are equal i.e.,

AB = DC.

To prove: Trapezium ABCD is cyclic.

Construction: Draw AM and DN so that they are perpendicular to BC.

Proof: In right triangles AMB and DNC,

∠AMB = ∠DNC = 90o

AB = DC [Given]

Since the perpendicular distance between the two parallel lines are same,

AM = DN

ΔAMB ≅ ΔDNC [By RHS congruence rule]

∠B = ∠C [CPCT]

And ∠1 = ∠2

∠BAD = ∠1+ 90

= ∠2 + 90

= ∠CDA

Now, in quadrilateral ABCD

∠B +∠C +∠CDA+∠BAD = 360

∠B +∠B +∠CDA+∠CDA = 360

2(∠B +∠CDA) = 360

∠B +∠CDA =180

We know,

If any pair of opposite angles of the quadrilateral is 180o, then the quadrilateral is cyclic.

Hence, the trapezium ABCD is cyclic.

Question 26: If P, Q and R are mid-points of sides BC, CA, and AB of a triangle and AD is perpendicular from point A on BC, prove that P, Q, R and D are concyclic.

Solution 26:

To prove: R, D, P and Q are concyclic.

Construction: We join RD, QD, PR and PQ.

RP is joining the mid-point of AB, i.e., R, and the mid-point of BC, i.e., P.

By using the midpoint theorem,

RP||AC

In a similar way,

PQ||AB.

Now, we get,

ARPQ is a parallelogram.

So, ∠RAQ = ∠RPQ [Opposite angles of a parallelogram]…(1)

ABD is a right-angled triangle while DR is a median,

RA = DR and ∠1 = ∠2 …(2)

Similarly, ∠3 = ∠4 …(3)

Now adding equations (2) and (3),

We get,

∠1+ ∠3 = ∠2 +∠4

⇒ ∠RDQ = ∠RAQ

∠RPQ [Proved earlier]

As ∠D and ∠P are subtended by RQ on the same side of it, we get the points R, D, P and Q as concyclic.

Hence proved that R, D, P and Q are concyclic.

Question 27:  ABCD is a parallelogram. A circle through the points A and B is drawn such that it is intersecting AD at P and BC at Q. Prove that P, Q, C, and D are concyclic.

Solution 27:

As per the question,

ABCD is a parallelogram.

A circle through points A and B is drawn such that it intersects AD at P and BC at Q.

To prove: P, Q, C and D are concyclic.

Construction: We join PQ.

We extend the side AP of the cyclic quadrilateral APQB to D.

External angle, ∠1= interior opposite angle, ∠B

BA||CD and BC cut them

∠B +∠C =180o

Sum of interior angles of the same side of the transversal = 180o

Or ∠1+∠C =180o

So, PDCQ is cyclic quadrilateral.

Therefore, the points P, Q, C and D are concyclic.

Question 28: Prove that angle bisector of an angle of the triangle and perpendicular bisector of the opposite side, if intersecting, they will be intersecting on the circumcircle of the triangle.

Solution 28:

As per the question,

ABC is a triangle, and l is a perpendicular bisector of BC.

To prove:

Angle bisector of ∠A and perpendicular bisector of BC intersecting on the circumcircle of ΔABC.

Proof:

We suppose the angle bisector of ∠A intersects the circumcircle of ΔABC at D.

Construction: We join BP and CP.

Since angles are equal in the same segment

We have ∠BAP = ∠BCP

We know,

AP is the bisector of ∠A.

Then,

∠BAP = ∠BCP = ½ ∠A …eq(1)

Similarly,

We have,

∠PAC = ∠PBC = ½ ∠A …e(2)

From equations (1) and (2),

We have

∠BCP = ∠PBC

We know,

If the angles subtended by the two chords of a circle at the centre are equal, then the chords are equal.

So,

BP = CP

Here, P is on the perpendicular bisector of BC.

Hence, angle bisector of ∠A and the perpendicular bisector of BC intersect on the circumcircle of ΔABC.

Question 27: Show that the line joining the point of intersection to the centre of the circle forms equal angles with the two equal chords of the circle.

Solution 27:

Given: AB and CD be any two equal chords of the circle with its centre O intersecting each other within the circle at point X.

OX is joined.

To prove: ∠OXM=∠OXN

Construction: Draw OM⊥AB and ON⊥CD

Proof: In right-angled triangles OMX and ONX,

∠OMX=∠ONX [Each 90∘]

OM=ON (Equal chords equidistant from the centre)

OX=OX (Common)

ΔOMX≅ΔONX (RHS rule of congruence)

∠OXM=∠OXN (By CPCT)

Question 28: In the figure, A, B, and C are the three points on a circle with centre O such that ∠BOC=30∘,∠AOB 60∘.∠BOC=30∘,∠AOB=60∘. If D is any point on the circle other than the arc ABC, find ∠ADC.

Solution 28:

∠AOC=∠AOB+∠BOC

⇒∠AOC= 60∘+30∘=90∘

Now

∠AOC=2∠ADC

Angled when subtended by an arc, at the centre of a circle is double the angle which is subtended by the same arc at any point in the remaining part of the circle

⇒∠ADC= 12∠AOC

⇒∠ADC= 12×90∘

=45∘

Question 29: In the figure,

∠PQR=100∘, where P, Q, and R are points on a circle with a centre  O. Find ∠OPR.

Solution 29: In the figure, Q is a point in the minor arc PQRˆ

∴mRPˆ=2∠PQR

⇒∠ROP=2∠PQR

⇒∠ROP=2×100∘=200∘

Now, mPRˆ+mRPˆ=360∘

⇒∠POR+∠ROP=360∘

⇒∠POR+200∘=360∘

⇒∠POR=360∘ -200∘=160∘…..(i)

Now, ΔOPR

ΔOPR is an isosceles triangle.

∴OP=OR (radii of circle)

⇒∠OPR=∠ORP (opposite angles to equal sides are equal)…..(ii)

Now in isosceles triangle OPR,

∠OPR+∠ORP+∠POR= 180∘

⇒∠OPR+∠ORP+ 160∘=180∘

⇒2∠OPR=180∘−160∘        

[Using(i) & (ii)]

⇒2∠OPR= 20∘

⇒∠OPR=10∘

Question 30: ABCD is a cyclic quadrilateral whose diagonals intersect at a point E.

∠DBC=70∘

∠BAC is 30∘ find ∠BCD. Further, if AB=BC, find ∠ECD.

Solution 30:  Here,

∠DBC=70∘ and ∠BAC=30∘

And ∠DAC=∠DBC=70∘ (Angles in same circle)

Now ABCD is a cyclic quadrilateral.

∴∠DAB+∠BCD=180∘ (Sum of opposite angles of a cyclic quadrilateral is supplementary)

⇒100∘+∠BCD=180∘

⇒∠BCD=80∘

Question 31: If the non-parallel sides of a trapezium are equal, can you prove that it is cyclic?

Solution 31: Given: A trapezium ABCD in which and AD=BC.

To prove: We are to prove that points A, B, C, and D are concyclic.

Construction: Draw DE.

Proof: Since DE||CB  and EB||DC.

∴EBCD is a parallelogram.

∴DE=CB and

∠DEB=∠DCB

Now AD=BC and DA=DE

⇒∠DAE=∠DEB

But ∠DEA+∠DEB=180∘

⇒∠DAE+∠DCB=180∘[∵∠DEA=∠DAE and ∠DEB=∠DCB]

⇒∠DAB+∠DCB=180∘

⇒∠A+∠C=180∘

Hence, ABCD is a cyclic trapezium.

Question 32: Two circles intersect at two points that are at B and C. Through B, two line segments ABD and PBQ are being drawn to intersect the circles at points A, D, P, and Q, respectively (see figure). Prove that ∠ACP=∠QCD

Solution 32: In the triangles ACD and QCP, 

∠A=∠P and ∠Q=∠D (Angles in the same segment)

∴∠ACD=∠QCP (Third angles) ……….(i)

Subtracting

∠PCD from both sides of eq. (i), we get,

∠ACD−∠PCD=∠QCP−∠PCD

⇒∠ACPO=∠QCD

Hence proved.

Question 33:  Prove that the line of centres of the two intersecting circles subtends equal angles at two points of intersection.

Solution 33: Let two circles with centres A and B intersect each other at points C and D.

We have to prove ∠ACB=∠ADB

Proof: In triangles ABC and ABD,

AC=AD=r

BC=BD=r

AB=AB (Common)

∴△ABC≅ΔABD (SSS rule of congruency)

⇒∠ACB=∠ADB [ByCPCT]

Question 34: Prove that a circle drawn with any side of a rhombus as diameter and is passing through the point of intersection of the diagonals.

Solution 34: Let ABCD be a rhombus whose diagonals AC and BD intersect each other at O.

We know that diagonals of the rhombus bisect and are perpendicular to each other.

∴∠AOB=90∘

And if we draw a circle with side AB as the diameter, it will surely pass through point O (the point intersection of diagonals) because then

∠AOB=90∘ will be the angle in a semi-circle.

Question 35: AB=DC and the diagonal AC and BD intersect at P in a cyclic quadrilateral. Prove that ΔPAB≅ΔPDC

Solution 35: In ΔPAB and ΔPDC 

AB=DC

∠ABP=∠DCP (Angle in the same segment)

∠PAB=∠PDC (Angle in the same segment)

ΔPAB≅ΔPDC (ASA criterion)

Question 36: Prove that ∠CAD=∠CBD if ABC and ADC are two right triangles with common hypotenuse AC.

Solution 36:

∠ADC=∠ABC=90∘[AC is the common hypotenuse of its ΔsADC and ABC ]

∠ADC+∠ABC=180∘

⇒ Quadrilateral ABCD is cyclic

Now, chord CD subtends ∠CAD and ∠CBD

∠CAD=∠CBD (Angle in the same segment)

Question 37: In the isosceles triangle ABC, AB=AC and B, C intersects the sides AB and AC at D and E, respectively. Prove that DE || BC.

Solution 37: In triangle ABC, given AB=AC

So ∠B=∠C (i)

Now, BCED forms a cyclic quadrilateral

∠ADC=∠C………(ii)

From (i) and (ii), we get

∠ADE=∠B

Hence DE || BC

Question 38: Prove that cyclic parallelogram is a rectangle.

Solution 38: We suppose ABCD be the given cyclic parallelogram

∠A+∠C=180∘………..(i)

∠A=∠C

(Opposite angles of a parallelogram are equal)  ……..(ii)

From (i) and (ii)

2∠A=180∘

∠A=90∘

So, ABCD is a parallelogram with one angle of 90∘ .

Hence, ∠ABCD is a rectangle.

Question 39: A line passing through the centre of a circle. Suppose it bisects the chords AB and CD of the circle. Prove that  AB||CD.

Solution 39: The line EF passes through the centre O and then bisects the chord AB at P and chord CD at Q.

∴P is the mid-point of AB and Q is the mid-point of CD.

However, the chord is parallel to the line that connects a chord’s midpoint to the circle’s centre.

OP⊥AB and OQ⊥CD

∠OPB=∠OQD=90∘

∠OPB+∠OQD=180∘

Hence,  AB||CD.

Question 40: Recall that the two circles are congruent if they have the same radii. Prove that the equal chords of the congruent circles subtend equal angles at their centres.

Solution 40:

Recall that a circle is made up of points that are all equally spaced apart from the centre. Therefore, two circles are only congruent when all of their points are equally distant from their centres.

In the second part of the question, it is given that AB = CD, i.e. two equal chords.

Now, we are to prove that angle AOB is equal to angle COD.

Proof:

Considering the triangles ΔAOB and ΔCOD,

OA = OC and OB = OD (Since they are radii of the circle)

AB = CD (As given in the question)

So, by SSS congruency, ΔAOB ≅ ΔCOD

∴ By CPCT, we have,

∠AOB = ∠COD. (Hence proved).

Question 41: Prove that if the chords of congruent circles subtend equal angles at their centres, then those chords are equal.

Solution 41:

Considering the following diagram-

It is given that ∠AOB = ∠COD, i.e. they are equal angles.

Now, we are to prove that the line segments AB and CD are equal, i.e. AB = CD.

Proof:

In triangles AOB and COD,

∠AOB = ∠COD (given in the question)

OA = OC and OB = OD (these are radii of the circle)

So, by SAS congruency, ΔAOB ≅ ΔCOD.

∴ By the rule of CPCT, we have

AB = CD. (Hence proved).

Question 42: Draw the different pairs of circles. How many such points does each pair have in common? What is the maximum number of possible common points?

Solution 42:

In these two circles, there is no point in common.

Here, only the point “P” is common.

Here also, P is the only common point.

Here, two common points are P and Q.

There is no point is common in the above circle.

Question 43: Suppose you are given a circle. Give a construction to find its centre.

Solution 43:

The steps of construction to find the centre of the circle are:

Step I: We draw a circle first.

Step II: We draw two chords AB and CD, in the circle.

Step III: We then draw the perpendicular bisectors of AB and CD.

Step IV: We connect the two perpendicular bisectors at a point. This intersecting point of the two perpendicular bisectors is the centre of the circle.

Question 44: If two circles intersect each other at two points, prove that their centres will lie on the perpendicular bisector of the common chord.

Solution 44:

It is given in the question that two circles intersect each other at points P and Q.

To prove:

OO’ is the perpendicular bisector of PQ.

Proof:

Triangle ΔPOO’ and ΔQOO’ are similar as established by SSS congruency 

OP = OQ and O’P = OQ (Since they are the radii)

OO’ = OO’ (It is a common side)

Thus, It can be said that ΔPOO’ ≅ ΔQOO’

∴ ∠POO’ = ∠QOO’ — (i)

Both the triangles ΔPOR and ΔQOR are similar by SAS congruency rule as

OP = OQ (Radii)

∠POR = ∠QOR (As ∠POO’ = ∠QOO’)

OR = OR (Common arm)

So, ΔPOR ≅ ΔQOR

Therefore, ∠PRO = ∠QRO

We also know that

∠PRO+∠QRO = 180°

Hence, ∠PRO = ∠QRO = 180°/2 = 90°

Thus, OO’ is the perpendicular bisector of PQ.

Question 45: Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.

Solution 45:

The common chord perpendicular bisector passes across the centres of both circles.

As the circles are intersecting at two points, we can construct the above figure as follows.

We consider AB as the common chord and O and O’ as the centres of the circles.

O’A = 5 cm

OA = 3 cm

OO’ = 4 cm [Distance between the centres is 4 cm]

We know that the smaller circle’s centre is inside, the larger circle because the radius of the larger circle is greater than the distance between its two centres.

The perpendicular bisector of AB is OO’

OA = OB = 3 cm

We know O is the midpoint of AB

AB = 3 cm + 3 cm = 6 cm

The length of the common chord is 6 cm

It is clear from the above that the common chord is the diameter of the smaller circle.

Question 46: If two equal chords of the circle intersecting within the circle, prove that the segment of one chord are equal to the corresponding segments of the other chord.

Solution 46:

Let AB and CD be the two equal chords (i.e. AB = CD). 

Given that AB and CD are intersecting at a point suppose E.

We are to prove that the line segments AE = DE and CE = BE

Steps of Construction:

Step 1: From the centre of the circle, we draw a perpendicular to AB, i.e. OM ⊥ AB

Step 2: Similarly, we draw ON ⊥ CD.

Step 3: We join OE.

Now, the diagram is as follows-

Proof:

From the diagram, 

OM bisects AB 

Thus, OM ⊥ AB

In the similar manner, ON bisects CD, and so, ON ⊥ CD

We know AB = CD. So,

AM = ND — (i)

and MB = CN — (ii)

Now, triangles ΔOME and ΔONE are similar because of RHS congruency since

∠OME = ∠ONE (They are perpendiculars)

OE = OE (It is the common side)

OM = ON (AB=CD as they are equidistant from the centre)

∴ ΔOME ≅ ΔONE

ME = EN (by CPCT) — (iii)

Now, from Eqs. (i) and (ii) we have,

AM+ME = ND+EN

So, AE = ED

Now from equations (ii) and (iii), we get,

MB-ME = CN-EN

Thus, EB = CE (Hence proved).

Question 47: If two equal chords of a circle are intersecting within the circle, prove that the line joining the point of intersection to the centre is making equal angles with the chords.

Solution 47:

 According to the question,

(i) AB and CD are two chords intersecting at point E.

(ii) PQ is diameter of the circle.

(iii) AB = CD.

To prove: ∠BEQ = ∠CEQ

For this, the following construction steps have to be followed:

Construction:

We draw two perpendiculars as OM ⊥ AB and ON ⊥ D. Also, we join OE. The constructed diagram is as follows:

Now, we consider the triangles ΔOEM and ΔOEN.

Here,

(i) OM = ON [Since equal chords are always equidistant from the centre]

(ii) OE = OE [common side]

(iii) ∠OME = ∠ONE [perpendiculars]

So, by RHS congruency Rule, ΔOEM ≅ ΔOEN.

Hence, by CPCT rule, ∠MEO = ∠NEO

∴ ∠BEQ = ∠CEQ (Hence proved).

Question 48:  If a line is intersecting two concentric circles (i.e., circles with the same centre) with centre O at A, B, C, and D, prove that AB = CD (see figure given below).

Solution 48:

First, we draw a line segment from O to AD such that OM ⊥ AD.

So OM has been bisecting AD since OM ⊥ AD.

Therefore, AM = MD — equation(i)

Also, since OM ⊥ BC, OM bisects BC.

Therefore, BM = MC — (ii)

From equation (i) and equation (ii),

AM-BM = MD-MC

∴ AB = CD

Question 49: Three girls, Reshma, Salma, and Mandip, are playing a game while standing on a circle with a radius of 5mwhich is drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m in each case, what will be the distance between Reshma and Mandip?

Solution 49:

We suppose the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively.

As per the question, we know that AB = BC = 6 cm.

So, the radius of  circle i.e. OA = 5cm

Now, we draw a perpendicular BM ⊥ AC.

Since AB = BC, ABC can be taken as an isosceles triangle. Then, M is mid-point of AC. BM is perpendicular bisector of AC, and it passes through the centre of the circle.

We suppose AM = y and

OM = x

So, BM will be = (5-x).

Now by using Pythagorean theorem in ΔOAM, we get,

OA2 = OM2 +AM2

⇒ 52 = x2 +y2 — (i)

Again, in ΔAMB, using the Pythagorean theorem,

AB2 = BM2 +AM2

⇒ 62 = (5-x)2+y2 — (ii)

Subtracting equation (i) from equation (ii), we have

36-25 = (5-x)2 +y2 -x2-y2

Solving this equation, we get the value of x as

x = 7/5

Putting the value of x in equation (i), we have

y2 +(49/25) = 25

⇒ y2 = 25 – (49/25)

Again solving it, we will get the value of y as

y = 24/5

Hence,

AC = 2×AM

= 2×y

= 2×(24/5) m

AC = 9.6 m

Therefore, the distance between Reshma and Mandip in the park is 9.6 m.

Question 50:  

A circular park of radius 20 m is situated in a colony. Three boys, Ankur, Syed, and David, are sitting at an equal distance on its boundary, each of whom has a toy telephone in their hands to talk to each other. Find the length of the string of each phone.

Solution 50:

First, we draw a diagram according to the given statements in the question. The diagram is represented as follows.

Here the positions of Ankur, Syed and David in the park are represented as A, B and C, respectively. Since they are all sitting at equal distances, the triangle ABC formed will be an equilateral triangle.

AD ⊥ BC is drawn. Now, AD is the median of ΔABC, which passes through the centre O.

Also, O is a centroid of the ΔABC. OA is the radius of the triangle.

OA = 2/3 AD

Let the side of a triangle a metre, then BD = a/2 m.

In ΔABD, using Pythagoras theorem,

AB2 = BD2+AD2

⇒ AD2 = AB2 -BD2

⇒ AD2 = a2 -(a/2)2

⇒ AD2 = 3a2/4

⇒ AD = √3a/2

OA = 2/3 AD

20 m = 2/3 × √3a/2

a = 20√3 m

Therefore, the length of the toy is 20√3 m.

Question 51:  In the figure given below, A, B, and C are the three points on a circle with the centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is any point on the circle other than the arc ABC, find ∠ADC.

Solution 51:

It is given that,

∠AOC = ∠AOB+∠BOC

So, ∠AOC = 60°+30°

∴ ∠AOC = 90°

We are aware that any angle an arc subtends at the centre of a circle is twice as large as any angle it subtends anywhere else on the rest of the circle.

So,

∠ADC = (½)∠AOC

= (½)× 90° = 45°

Question 52: A chord of a given circle is equal to the radius of the circate. Find the angle subtended by the chord at any point on the minor arc and also at any point on the major arc.

Solution 52:

Here, the chord AB is equal to the radius of the circle. In the above-represented diagram, OA and OB are two radii of the circle.

Now, considering the ΔOAB. Here,

AB = OA = OB = radius of circle.

So, we can say that ΔOAB has all sides equal and therefore, it is an equilateral triangle.

∴ ∠AOC = 60°

And, ∠ACB = ½ ∠AOB

Thus, ∠ACB = ½ × 60° = 30°

Now, as ACBD is a cyclic quadrilateral,

∠ADB +∠ACB = 180° (opposite angles of a cyclic quadrilateral)

So, ∠ADB = 180°-30° = 150°

So, the angle subtended by the chord at any point on the minor arc and also at any point on the major arc are 150° and 30°, respectively.

Question 53: In the figure given below,, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution 53:

An arc’s angle at the centre of the circle is twice as large as its angle at any other point on the remaining portion of the circle.

So, the reflex ∠POR = 2×∠PQR

As the values ∠PQR as 100°

So, ∠POR = 2×100° = 200°

∴ ∠POR = 360°-200° = 160°

Now, in ΔOPR,

OP and OR are radii of the circle

So, OP = OR

and ∠OPR = ∠ORP

Now, we know that the sum of the angles in a triangle is equal to 180 degrees

So,

∠POR+∠OPR+∠ORP = 180°

∠OPR+∠OPR = 180°-160°

As ∠OPR = ∠ORP

2∠OPR = 20°

Thus, ∠OPR = 10°

Question 54:  In figure given below, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution 54:

As we know, the angles in the segment of the circle are equal, so,

∠BAC = ∠BDC

In ΔABC, the sum of all the interior angles is 180°

So, ∠ABC+∠BAC+∠ACB = 180°

Now, by putting the values,

∠BAC = 180°-69°-31°

So, ∠BAC = 80°

∴ ∠BDC = 80°

Question 55:  In the figure given below, A, B, C and D are the four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find BAC.

Solution 55:

We know the angles in the segment of the circle are equal.

So,

∠ BAC = ∠ CDE

Now, using the exterior angles property of the triangle

In ΔCDE, we get,

∠ CEB = ∠ CDE+∠ DCE

We know,

 ∠ DCE is equal to 20°

So, ∠ CDE = 110°

∠ BAC and ∠ CDE are equal

∴ ∠ BAC = 110°

Question 56: ABCD is a cyclic quadrilateral with diagonals intersecting at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Solution 56:

Considering the following diagram.

Considering the chord CD,

We know angles that are in the same segment are equal.

So, ∠ CBD = ∠ CAD

∴ ∠ CAD = 70°

Now, ∠ BAD is equal to the sum of angles BAC and CAD.

So, ∠ BAD = ∠ BAC+∠ CAD

= 30°+70°

∴ ∠ BAD = 100°

Again we know the opposite angles of a cyclic quadrilateral adds up to 180 degrees.

So,

∠ BCD+∠ BAD = 180°

Again, ∠ BAD = 100°

So, ∠ BCD = 80°

Now considering the ΔABC.

Here, it is given that AB = BC

Also, ∠ BCA = ∠ CAB (i.e., angles opposite to equal sides of a triangle)

∠ BCA = 30°

also, ∠ BCD = 80°

∠ BCA +∠ ACD = 80°

Therefore, ∠ ACD = 50° and ∠ ECD = 50°

Question 57:  If the diagonals of a cyclic quadrilateral are diameters of the circle when drawn through the vertices of the quadrilateral, prove that it is a rectangle.

Solution 57:

We draw a cyclic quadrilateral ABCD inside a circle with its centre O such that its diagonal AC and BD are the two diameters of the circle.

We know the angles in the semi-circle are equal.

So, ∠ ABC = ∠ BCD = ∠ CDA = ∠ DAB = 90°

Since each internal angle is 90°, it can be said that the quadrilateral ABCD is a rectangle.

Question 58: Two circles are intersecting at two points, B and C. Through B, two line segments ABD and PBQ are drawn such that it intersect the circles at A, D and P, Q, respectively (see figure given below). Prove that ∠ ACP = ∠ QCD.

Solution 58:

Construction:

We join the chords AP and DQ.

For chord AP, we know angles in the same segment are equal.

So, ∠ PBA = ∠ ACP — eq(i)

Similarly, for chord DQ,

∠ DBQ = ∠ QCD — eq(ii)

We know that ABD and PBQ are two line segments intersecting at point B.

At point B, the vertically opposite angles will be equal.

∴ ∠ PBA = ∠ DBQ — eq(iii)

Thus from equation (i), equation (ii) and equation (iii), we get,

∠ ACP = ∠ QCD

Question 59: If circles are being drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles will be lying on the third side.

Solution 59:

First, we draw a triangle ABC, and then two circles having diameters as AB and AC, respectively.

To prove that D lies on BC and BDC is a straight line.

Proof:

We know angle in the semi-circle are equal.

So, ∠ ADB = ∠ ADC = 90°

Thus, ∠ ADB+∠ ADC = 180°

∴ ∠ BDC is a straight line.

So, it can be said that D will lie on the line BC.

Question 60: ABC and ADC are the two right triangles having a common hypotenuse AC. Prove that ∠ CAD = ∠CBD.

Solution 60:

We know that AC is a common hypotenuse and ∠ B = ∠ D = 90°.

To prove that ∠ CAD = ∠ CBD

Since ∠ ABC and ∠ ADC are 90°, they lie in the semi-circle.

Thus, triangles ABC and ADC are in the semi-circle, and the points A, B, C and D are concyclic.

Hence, CD is the chord of a circle with its centre O.

We know the angles in the same segment of the circle are also equal.

∴ ∠ CAD = ∠ CBD

Question 61:  Prove that the centre lines of two intersecting circles subtends equal angles at the two points of intersection.

Solution 61:

Consider the following diagram

In ΔPOO’ and ΔQOO’

OP = OQ          (Radius of circle 1)

O’P = O’Q        (Radius of circle 2)

OO’ = OO’        (Common arm)

So, by SSS congruency, ΔPOO’ ≅ ΔQOO’

Thus, ∠OPO’ = ∠OQO’ (proved).

Question 62: Two chords AB and CD of lengths 5 cm and 11 cm, respectively, of a given circle, are parallel to each other and are on the opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.

Solution 62:

Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.

We know that AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2 = 5/2

In the similar manner, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6−x.

Considering ΔMOB,

OB2 = OM2+MB2

Or,

OB2 = 36 + X2 – 12x + 254       ……..(1)

Consider ΔNOD,

OD2 = ON2 + ND2

Or

OD2 = X2 + 1214       …(2)

OB = OD (radii)

From equation 1 and equation 2, we have,

36 + X2 – 12x + 254 = X2 + 1214

12x = 36 + 254  – 1214

= 144 + 25 -1214

12x = 484 = 12

x= 1

Now, from equation (2), we have,

OD2= 12 +(121/4)

Or OD = (5/2)×√5 cm

Question 63: The lengths of the two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a 4 cm distance from the centre, what will be the distance of the other chord from the centre?

Solution 63:

Considering the following diagram

Here AB and CD are two parallel chords. Now, we join OB and OD.

Distance of smaller chord AB from the centre of the circle = 4 cm

Thus, OM = 4 cm

MB = AB/2 = 3 cm

Considering ΔOMB

OB2 = OM2+MB2

Or, OB = 5cm

Now, considering ΔOND,

OB = OD = 5 (since radii)

ND = CD/2 = 4 cm

Now, OD2= ON2+ND2

Or, ON = 3 cm.

Question 64: Suppose the vertex of an angle ABC be located outside a circle and let the sides of the angle be intersecting equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference between the angles when subtended by the chords AC and DE at the centre.

Solution 64:

Considering the diagram,

Here AD = CE

We know any exterior angle of a triangle equals the sum of the interior opposite angles.

So,

∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-eq(i)

DE subtends ∠DOE at the centre while ∠DAE subtends in the remaining part of the circle.

So,

∠DAE = (½)∠DOE ——————-eq(ii)

In the similar way, ∠AEC = (½)∠AOC  ——————-eq(iii)

Now, from equation (i), (ii), and (iii), we have,

(½)∠DOE = ∠ABC+(½)∠AOC

Or, ∠ABC = (½)[∠DOE-∠AOC]

Hence proved.

Question 65: Prove that the circle, when drawn with any side of the rhombus as diameter, passes through the point of intersection of the diagonals.

Solution 65:

To prove: A circle drawn with Q as the centre will pass through A, B and O (i.e. QA = QB = QO)

Since we know all sides of a rhombus are equal, therefore

AB = DC

Now, multiplying (½) on both sides

(½)AB = (½)DC,

Therefore, AQ = DP

BQ = DP

As Q is the midpoint of AB,

AQ= BQ

In the same manner,

RA = SB

Again, when PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO, we have,

QA = QB = QO (hence proved).

Question 66: ABCD is a parallelogram. The circle through A, B, and C is intersecting CD (if produced) at E. Prove that AE = AD.

Solution 66:

Here, ABCE is a given cyclic quadrilateral. In any cyclic quadrilateral, the sum of opposite angles is 180°.

So, ∠AEC+∠CBA = 180°

When ∠AEC and ∠AED are linear pairs,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … eq(1)

We know in a parallelogram, the opposite angles are equal.

So, ∠ADE = ∠CBA … eq(2)

Now, from equations (1) and (2), we have,

∠AED = ∠ADE

Now, AD and AE are opposite angles to equal sides of a triangle,

∴ AD = AE (proved).

Question 67:  AC and BD are the chords of a circle which bisects each other. Prove that (i) AC and BD are both diameters; (ii) ABCD is the rectangle.

Solution 67:

Here the chords AB and CD intersect each other at O.

Considering ΔAOB and ΔCOD,

∠AOB = ∠COD (vertically opposite angles)

OB = OD (Given)

OA = OC (Given)

So, by SAS congruency rule, ΔAOB ≅ ΔCOD

Also, AB = CD (By CPCT)

In the similar manner, ΔAOD ≅ ΔCOB

Or, AD = CB (By CPCT)

In quadrilateral ACBD, the opposite sides are equal.

Hence, ACBD is a parallelogram.

We know the opposite angles of a parallelogram are equal.

So, ∠A = ∠C

Also, as ABCD is a cyclic quadrilateral,

∠A+∠C = 180°

⇒∠A+∠A = 180°

Or, ∠A = 90°

As ACBD is a parallelogram whose one of the interior angles is 90°, so, it is a rectangle.

∠A is the angle subtended by the chord BD. And as ∠A = 90°, thus, BD should be diameter of the circle. Similarly, AC is diameter of the circle.

Question 68: The bisectors of the angles A, B, and C of triangle ABC intersects its circumcircle at D, E and F, respectively. Prove that the angles of triangle DEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.

Solution 68:

Considering the following diagram,

Here, ABC is a triangle inscribed in a circle with its centre O and the bisectors of ∠A, ∠B and ∠C intersecting the circumcircle at D, E and F, respectively.

Now, we join DE, EF and FD

As angles are equal in the same segment, therefore,

∠EDA = ∠FCA ————-eq(i)

∠FDA = ∠EBA ————-eq(i)

By adding equations (i) and (ii), we have,

∠FDA+∠EDA = ∠FCA+∠EBA

Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B

We know that ∠A +∠B+∠C = 180°

Thus, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]

∠FDE = [90-(∠A/2)]

In a similar way,

∠FED = [90° -(∠B/2)] °

And,

∠EFD = [90° -(∠C/2)] °

Question 69: Two congruent circles intersecting each other at points A and B. Through point A, any line segment PAQ is drawn so that P and Q lie on the two circles. Prove that BP = BQ.

Solution 69:

The diagram will be as follows:

Here, ∠APB = ∠AQB (as AB is a common chord in both the congruent circles.)

Now, considering ΔBPQ,

∠APB = ∠AQB

So, the opposite angles to equal sides of a triangle.

∴ BQ = BP

Question 70:  In triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersects, prove that they are intersecting on the circumcircle of the triangle ABC.

Solution 70:

Considering this diagram

 

Here, we join BE and CE.

Now, since AE is bisector of ∠BAC,

∠BAE = ∠CAE

Also,

 arc BE = arc EC

Thus, chord BE = chord EC

Now, considering triangles ΔBDE and ΔCDE,

DE = DE     (common side)

BD = CD     (given in the question)

BE = CE      (proved earlier)

So, by SSS congruency, ΔBDE ≅ ΔCDE.

Thus, ∴∠BDE = ∠CDE

We know, ∠BDE = ∠CDE = 180°

Or, ∠BDE = ∠CDE = 90°

∴ DE ⊥ BC (hence proved).

Benefits of Solving Important Questions Class 9 Mathematics Chapter 10

The following is a summary of the benefits of solving questions from our Class 9 Mathematics Chapter 10 important questions.

• Our Important Questions Class 9 Mathematics Chapter 10 question bank is based on the most recent CBSE syllabus and is written in the manner of an NCERT exam. In order to expose students to the actual board test patterns, it, therefore, includes questions in a number of formats, such as multiple-choice questions (MCQs), fill-in-the-blank questions, and short-form and long-form responses. Students are encouraged to perform better on exam day and score better marks.
• The solutions have been produced by Mathematics teachers with years of experience. The expert members continually review and enhance the resources. Students can therefore rely on our solutions with certainty as they all adhere to the most recent CBSE syllabus.
• By practising the different problems in the Important Questions Class 9 Mathematics Chapter 10, students can thoroughly study the concepts covered in the chapter. Before their final exams, they will have the chance to review the subject once more and identify and work on any doubts. It is important to stick to a regular study schedule if you wish to succeed.

All of the important topics are sufficiently covered with explanations in the answers to all of the questions in our Important Questions Class 9 Mathematics Chapter 10.

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